Simplify; express your answer in exponential form. Assume $n\neq 0, z\neq 0$. $\dfrac{{(n^{2})^{-5}}}{{(n^{-4}z^{2})^{-5}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{2}}$ to the exponent ${-5}$ . Now ${2 \times -5 = -10}$ , so ${(n^{2})^{-5} = n^{-10}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-4}z^{2})^{-5} = (n^{-4})^{-5}(z^{2})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{2})^{-5}}}{{(n^{-4}z^{2})^{-5}}} = \dfrac{{n^{-10}}}{{n^{20}z^{-10}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-10}}}{{n^{20}z^{-10}}} = \dfrac{{n^{-10}}}{{n^{20}}} \cdot \dfrac{{1}}{{z^{-10}}} = n^{{-10} - {20}} \cdot z^{- {(-10)}} = n^{-30}z^{10}$.